Chapter 9 Testing a Claim 9.1 Significance Tests: The Basics 9.2 Tests about a Population Proportion 9.3 Tests about a Population Mean

Section 9.2 Tests About a Population Proportion Learning Objectives After this section, you should be able to… CHECK conditions for carrying out a test about a population proportion. CONDUCT a significance test about a population proportion. CONSTRUCT a confidence interval to draw a conclusion for a two- sided test about a population proportion.

Two-Sided Tests According to the Centers for Disease Control and Prevention (CDC) Web site, 50% of high school students have never smoked a cigarette. Tyson wonders whether this national result holds true in his large, urban high school. For his AP Statistics class project, Tyson surveys an SRS of 150 students from his school. He gets responses from all 150 students, and 90 say that they have never smoked a cigarette. What should Tyson conclude? Give appropriate evidence to support your answer. State: We want to perform a test at the a = 0.05 significance level of H0: p = 0.50 Ha: p ≠ 0.50 where p = the actual proportion of students in Tyson’s school who would say they have never smoked cigarettes. Plan: If the conditions are met, we will do a one-sample z test for the population proportion p. Random: Tyson surveyed an SRS of 150 students from his school. Normal: Assuming H0: p = 0.50 is true, np0 = 150(0.50) = 75 and n(1 – p0) = 150(0.50) = 75 which are both ≥ 10. We can assume a Normal distribution. Independent: Since we are sampling w/o replacement, we need to check the 10% condition. It seems reasonable to assume that there are at least 10(150) = 1500 students at this school.

Two-Sided Tests According to the Centers for Disease Control and Prevention (CDC) Web site, 50% of high school students have never smoked a cigarette. Tyson wonders whether this national result holds true in his large, urban high school. For his AP Statistics class project, Tyson surveys an SRS of 150 students from his school. He gets responses from all 150 students, and 90 say that they have never smoked a cigarette. What should Tyson conclude? Give appropriate evidence to support your answer. Do: The sample proportion is Test Statistic: Standardized P-Value: To compute this P-value, we find the area in one tail and double it. 2P(z ≥ 2.45) = 2(0.0071) = 0.0142 normalcdf(2.45, E99, 0, 1) Conclude: Since our P-value, 0.0142, is less than the chosen significance level of a = 0.05, we reject H0 and conclude the proportion of students at Tysons school who -3 -2 -1 1 2 3 say they have never smoked differs from the national z = -2.45 z = 2.45 average of 0.50.

Why Confidence Intervals Give More Information The result of a significance test is basically a decision to reject H0 or fail to reject H0. When we reject H0, we’re left wondering what the actual proportion p might be. A confidence interval might shed some light on this issue. Tyson found that 90 of an SRS of 150 students said that they had never smoked a cigarette. Before we construct a confidence interval for the population proportion p, we should check that both the number of successes and failures are at least 10. For a confidence interval, we use . are both ≥ 10. Our 95% confidence interval is: (0.522, 0.678) 0.50 is not in this interval – reject. We are 95% confident the interval from 0.522 to 0.678 captures the true proportion of students at Tyson’s school who say they never smoked.

Confidence Intervals and Two-Sided Tests There is a link between confidence intervals and two-sided tests. The 95% confidence interval gives an approximate range of p0’s that would not be rejected by a two-sided test at the α = 0.05 significance level. The link isn’t perfect because the standard error used for the confidence interval is based on p-hat, while the denominator of the test statistic is based on the value p0 from the null hypothesis. Pencils Down A two-sided test at significance level of a = 5% and a 95% confidence interval give similar information about the population parameter. If the sample proportion falls in the “fail to reject H0” region, like the green value in the figure, the resulting 95% confidence interval would include p0. In that case, both the significance test and the confidence interval would be unable to rule out p0 as a plausible parameter value. However, if the sample proportion falls in the “reject H0” region, the resulting 95% confidence interval would not include p0. In that case, both the significance test and the confidence interval would provide evidence that p0 is not the parameter value.

Example: Benford’s law and fraud When the accounting firm AJL and Associates audits a company’s financial records for fraud, they often use a test based on Benford’s law. Benford’s law states that the distribution of first digits in many real-life sources of data is not uniform. In fact, when there is no fraud, about 30.1% of the numbers in financial records begin with the digit 1. However, if the proportion of first digits that are 1 is significantly different from 0.301 in a random sample of records, AJL and Associates does a much more thorough investigation of the company. Suppose that a random sample of 300 expenses from a company’s financial records results in only 68 expenses that begin with the digit 1. Should AJL and Associates do a more thorough investigation of this company? State: We want to perform a test at the a = 0.05 significance level of H0: p = 0.301 Ha: p ≠ 0.301 where p = the true proportion of expenses that begin with the digit 1. Plan: If the conditions are met, we will do a one-sample z test for p. Random: A random sample of expenses was selected. Normal: Assuming H0 is true, np0 = 300(0.301) = 90.35 and n(1 – p0) = 300(0.699) = 209.7 which are both ≥ 10. We can assume a Normal distribution. Independent: Since we are sampling w/o replacement, we need to check the 10% condition. It seems reasonable to assume that there are at least 10(300) = 3000 company expenses.

Example: Benford’s law and fraud When the accounting firm AJL and Associates audits a company’s financial records for fraud, they often use a test based on Benford’s law. Benford’s law states that the distribution of first digits in many real-life sources of data is not uniform. In fact, when there is no fraud, about 30.1% of the numbers in financial records begin with the digit 1. However, if the proportion of first digits that are 1 is significantly different from 0.301 in a random sample of records, AJL and Associates does a much more thorough investigation of the company. Suppose that a random sample of 300 expenses from a company’s financial records results in only 68 expenses that begin with the digit 1. Should AJL and Associates do a more thorough investigation of this company? Do: The sample proportion is Test Statistic: Standardized P-Value: To compute this P-value, we find the area in one tail and double it. 2P(z ≤ -2.79) = 2(0.0026) = 0.0052 normalcdf(-E99, -2.79, 0, 1) Conclude: Since our P-value, 0.0052, is less than the chosen significance level of a = 0.05, we reject H0 and conclude the proportion of expenses that have a first -3 -2 -1 1 2 3 digit of 1 is not 0.301. Therefore AJL and Associates z = -2.79 z = 2.79 should investigate for fraud.

Example: Benford’s law and fraud When the accounting firm AJL and Associates audits a company’s financial records for fraud, they often use a test based on Benford’s law. Benford’s law states that the distribution of first digits in many real-life sources of data is not uniform. In fact, when there is no fraud, about 30.1% of the numbers in financial records begin with the digit 1. However, if the proportion of first digits that are 1 is significantly different from 0.301 in a random sample of records, AJL and Associates does a much more thorough investigation of the company. Suppose that a random sample of 300 expenses from a company’s financial records results in only 68 expenses that begin with the digit 1. Should AJL and Associates do a more thorough investigation of this company? Find and interpret a confidence interval for the true proportion of expenses that begin with the digit 1. State: We want to estimate p = the true proportion of expenses that begin with the digit 1 at the 95% confidence level. Plan: If the conditions are met, we will do a one-sample z interval for p. Random: A random sample of expenses was selected. Normal: and are both ≥ 10. We can assume a Normal distribution. Independent: The 10% condition is satisfied 10(300) = 3000.

Example: Benford’s law and fraud When the accounting firm AJL and Associates audits a company’s financial records for fraud, they often use a test based on Benford’s law. Benford’s law states that the distribution of first digits in many real-life sources of data is not uniform. In fact, when there is no fraud, about 30.1% of the numbers in financial records begin with the digit 1. However, if the proportion of first digits that are 1 is significantly different from 0.301 in a random sample of records, AJL and Associates does a much more thorough investigation of the company. Suppose that a random sample of 300 expenses from a company’s financial records results in only 68 expenses that begin with the digit 1. Should AJL and Associates do a more thorough investigation of this company? Find and interpret a confidence interval for the true proportion of expenses that begin with the digit 1. Do: Conclude: We are 95% confident that the interval from 0.180 to 0.274 captures the true proportion of expenses that begin with the digit 1.

Example: Benford’s law and fraud When the accounting firm AJL and Associates audits a company’s financial records for fraud, they often use a test based on Benford’s law. Benford’s law states that the distribution of first digits in many real-life sources of data is not uniform. In fact, when there is no fraud, about 30.1% of the numbers in financial records begin with the digit 1. However, if the proportion of first digits that are 1 is significantly different from 0.301 in a random sample of records, AJL and Associates does a much more thorough investigation of the company. Suppose that a random sample of 300 expenses from a company’s financial records results in only 68 expenses that begin with the digit 1. Should AJL and Associates do a more thorough investigation of this company? Use the confidence interval to decide whether this company should be investigated for fraud. Since 0.301 is not in our interval, 0.301 is not a plausible value for the true proportion of expenses that begin with the digit 1. Thus, this company should be investigated for fraud.

Section 9.2 Tests About a Population Proportion Summary In this section, we learned that… As with confidence intervals, you should verify that the three conditions— Random, Normal, and Independent—are met before you carry out a significance test. Significance tests for H0 : p = p0 are based on the test statistic with P-values calculated from the standard Normal distribution. The one-sample z test for a proportion is approximately correct when (1) the data were produced by random sampling or random assignment; (2) the population is at least 10 times as large as the sample; and (3) the sample is large enough to satisfy np0 ≥ 10 and n(1 - p0) ≥ 10 (that is, the expected numbers of successes and failures are both at least 10).

Section 9.2 Tests About a Population Proportion Summary In this section, we learned that… Follow the four-step process when you carry out a significance test: STATE: What hypotheses do you want to test, and at what significance level? Define any parameters you use. PLAN: Choose the appropriate inference method. Check conditions. DO: If the conditions are met, perform calculations. • Compute the test statistic. • Find the P-value. CONCLUDE: Interpret the results of your test in the context of the problem. Confidence intervals provide additional information that significance tests do not—namely, a range of plausible values for the true population parameter p. A two-sided test of H0 : p = p0 at significance level α gives roughly the same conclusion as a 100(1 – α)% confidence interval.

Looking Ahead… In the next Section… We’ll learn how to test a claim about a population mean. We’ll learn about Carrying out a significance test The one-sample t test for a mean Two-sided tests and confidence intervals Paired data and one-sample t procedures